Notes
three squares in a circle ii solution

Solution to the Three Squares in a Circle II Puzzle

Solution by Pythagoras' Theorem, Difference of Two Squares, and Area of a Circle

In the diagram above, $O$ is the centre of the circle and $E B$ is perpendicular to the sides of the squares at $E$ and $B$. The squares have side lengths $6$, $12$, and $18$. Therefore, $E B$ has length $36$.

Let $r$ be the radius of the circle. Let $x$ be the length of $O E$ and $y$ the length of $O B$, so $x + y = 36$. Applying Pythagoras' theorem to triangle $O E D$ shows that:

and applying it to triangle $O B C$ shows that:

Subtracting these equations shows that:

Since $x + y = 36$, and by difference of two squares $x^2 - y^2 = (x + y)(x - y)$ this means that $x - y = 2$. So $x = 19$ and $y = 17$.

Then $r^2 = 3^2 + 19^2 = 370$.

The area of the circle is therefore $370\pi$.